Question: $\int (-2 x^4 -4 x^2 -5)\,dx=$ $+C$
Explanation: We can use the sum rule and the constant multiple rule for indefinite integrals: $\begin{aligned} &\int [f(x)+g(x)]dx=\int f(x)\,dx+\int g(x)\,dx \\\\\\ &\int k\cdot f(x)= k\cdot\int f(x)\,dx \end{aligned}$ Using the sum and the constant multiple rules, we can rewrite our integral as follows: $\int (-2 x^4 -4 x^2 -5)\,dx= -2\int x^4\,dx -4\int x^2\,dx -5\int 1\,dx$ Now we can find each indefinite integral using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ Note: we can only use the reverse power rule because $n \neq -1$. $\begin{aligned} &\phantom{=}\int (-2 x^4 -4 x^2 -5)\,dx \\\\ &= -2\int x^4\,dx -4\int x^2\,dx -5\int 1\,dx \\\\ &=-2 \dfrac{x^5}{5} -4\dfrac{x^3}{3} -5\dfrac{x^1}{1}+C \\\\ &=-\dfrac{2}{5} x^5 -\dfrac{4}{3} x^3 -5 x+C \end{aligned}$ In conclusion, $\int (-2 x^4 -4 x^2 -5)\,dx=-\dfrac{2}{5} x^5 -\dfrac{4}{3} x^3 -5 x+C$